Which equation will solve the following word problem? News Center 13 has just released tomorrow's weather forecast. The high tem
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Answer: approximately 5148 feet directly north of point A
This is equivalent to 0.975 miles
==============================================================
Explanation:
Check out figure 1 (attached image below) to see how the drawing is set up.
Points A and B are separated by 2 miles, or 10560 feet. Use the conversion factor 1 mile = 5280 feet.
Because sound travels at roughly 1100 feet per second (assuming all of the conditions are right), this means that after x seconds, the sound wave has traveled a total distance of 1100x feet. This is the distance from A to C. Point C is the firework location.
The distance from B to C is 1100(x+6) feet because it takes 6 more seconds for the soundwave to travel from C to B, compared from C to A.
Note that triangle ABC is a right triangle. The 90 degree angle is at point A.
We can use the pythagorean theorem to solve for x
See figure 2 in the attached images to see the steps on solving for x. The numbers get quite big, though in the end x is fairly small. We end up with x = 4.68, which is approximate due to the 1100 figure being approximate.
This means the sound wave takes about 4.68 seconds to go from point C to point A.
With this x value, we can compute the expression 1100x to get
1100*x = 1100*4.68 = 5,148 feet
To convert this to miles, divide by 5280
(5148)/(5280) = 0.975
Therefore,
5148 feet = 0.975 miles
Proof of the Law of Sines
The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)
a
sin A
=
b
sin B
=
c
sin C
For more see Law of Sines.
Acute triangles
Draw the altitude h from the vertex A of the triangle
From the definition of the sine function
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Repeat the above, this time with the altitude drawn from point B
Using a similar method it can be shown that in this case
c
sin C
=
a
sin A
Combining (4) and (5) :
a
sin A
=
b
sin B
=
c
sin C
- Q.E.D
Obtuse Triangles
The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.
First the interior altitude. This is the same as the proof for acute triangles above.
Draw the altitude h from the vertex A of the triangle
sin B =
h
c
a n d sin C =
h
b
or
h = c sin B a n d h = b sin C
Since they are both equal to h
c sin B = b sin C
Dividing through by sinB and then sinC
c
sin C
=
b
sin B
Draw the second altitude h from B. This requires extending the side b:
The angles BAC and BAK are supplementary, so the sine of both are the same.
(see Supplementary angles trig identities)
Angle A is BAC, so
sin A =
h
c
or
h = c sin A
In the larger triangle CBK
sin C =
h
a
or
h = a sin C
From (6) and (7) since they are both equal to h
c sin A = a sin C
Dividing through by sinA then sinC:
a
sin A
=
c
sin C
Combining (4) and (9):
a
sin A
=
b
sin B
=
c
sin C
Answer:
The width and the length of the pool are 12 ft and 24 ft respectively.
Step-by-step explanation:
The length (L) of the rectangular swimming pool is twice its wide (W):
Also, the area of the walkway of 2 feet wide is 448:
Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).
The total area is related to the pool area and the walkway area as follows:
(1)
The area of the pool is given by:
(2)
And the area of the walkway is:
(3)
Where the length of the bigger rectangle is related to the lower rectangle as follows:
(4)
By entering equations (4), (3), and (2) into equation (1) we have:
By solving the above quadratic equation we have:
W₁ = 12 ft
Hence, the width of the pool is 12 feet, and the length is:
Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.
I hope it helps you!